Question

The concentration of 1L of \( CaCO_3 \) solution is 1000 ppm. What is its concentration in mol \( L^{-1} \)?
(Ca = 40 u, O = 16 u, C = 12 u)

• A. \( 10^{-3} \)
• B. \( 10^{-1} \)
• C. \( 10^{-4} \)
• D. \( 10^{-2} \)

Hint:

To convert ppm to molarity, use the relation: \[ \text{Molarity} = \frac{\text{ppm value}}{\text{Molar mass} \times 1000} \]

Answer 1

The Correct Option is D

We are given that the concentration of \( 1 \, \text{L} \) of \( \text{CaCO}_3 \) solution is \( 1000 \) ppm. We need to find its concentration in mol/L. We know that: \begin{itemize} \item \( \text{Ca} = 40 \, \mu \), \item \( \text{C} = 12 \, \mu \), \item \( 1 \, \text{L} \, \text{CaCO}_3 \) concentration = 1000 ppm. \end{itemize} We can calculate the molar concentration by converting ppm to mol/L using the molar mass of \( \text{CaCO}_3 \). The molecular weight of \( \text{CaCO}_3 \) is: \[ \text{Molar Mass of } \text{CaCO}_3 = \text{Ca} + \text{C} + 3 \times \text{O} \] \[ \text{Molar Mass of } \text{CaCO}_3 = 40 + 12 + 3 \times 16 = 40 + 12 + 48 = 100 \, \text{g/mol} \] Now, to convert ppm to mol/L: \begin{itemize} \item \( 1 \, \text{ppm} = 1 \, \text{mg/L} \), \item \( 1000 \, \text{ppm} = 1000 \, \text{mg/L} \), \item \( 1000 \, \text{mg/L} = \frac{1000}{100} = 10 \, \text{mol/L} \). \end{itemize} Thus, the concentration of the solution in mol/L is: \[ 10^{-2} \, \text{mol/L} \]